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Intersection between 2 sets means the elements that are common to both those sets. I read that at Wikipedia.

$\varnothing$ is the subset of every set but not an element of all the sets.

If $\varnothing$ is not an element of every set, then why is A intersection A' = null?

Edit: I just started reading sets. I had this doubt and I asked it. Sorry I might have done something wrong but please don't be angry at me.

Suppose A = {1,3,5} and U = {1,2,3,4,5} So A' = {2,4}. So why will A intersection A' = null, when null is not an element of any of the two sets (A and A')?

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    $\begingroup$ This question barely makes sense; what is $A^\prime$? If what is an element of every set? $\endgroup$ – Edward Evans Apr 4 '18 at 9:41
  • $\begingroup$ For the proof that $\emptyset$ is subset of every set, see the post : empty-set-subsets-and-vacuous-truths. $\endgroup$ – Mauro ALLEGRANZA Apr 4 '18 at 9:54
  • $\begingroup$ What you call "null" is the empty set : it is not listed between the elements of $A = \{ 1,3,5 \}$ and thus $\emptyset \notin A$. $\endgroup$ – Mauro ALLEGRANZA Apr 4 '18 at 10:49
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    $\begingroup$ Two sets are equal exactly when they have the same elements. $\emptyset$ is empty: i.e. it has no elements. We have shown that $A \cap A'$ has no elements (because there are no common elements to a set and its complement). Thus, the two sets: $A \cap A'$ and $\emptyset$ are equal, because both have no elements. $\endgroup$ – Mauro ALLEGRANZA Apr 4 '18 at 12:17
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    $\begingroup$ @MauroALLEGRANZA thank you. thank you. I understand now. Thank you. Sorry but should I delete this question or let it be? Or can you post that as an answer so that I can ✅ it? $\endgroup$ – Ram Keswani Apr 4 '18 at 12:22
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See the definition of Complement (of a set), denoted : $A^c,\overline A,A′$.

If $A′$ is the complement of the set $A$, obviously $A \cap A′= \emptyset$, because from $x ∈ A \cap A′$ we have that $x∈A$ and $x∈A′$.

The last one is equivalent to $x∉A$, and thus we have bot $x∈A$ and $x∉A$ : contradicition !

Thus there are no common elements to $A$ and $A'$, i.e. their intersection $A \cap A'$ is empty.

Two sets are equal exactly when they have the same elements.

$\emptyset$ is empty: i.e. it has no elements.

We have shown that $A \cap A′$ has no elements (because there are no common elements to a set and its complement).

Thus, the two sets: $A \cap A′$ and $\emptyset$ are equal, because both have no elements.

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I don't have enough reputation yet, otherwise I'd add this as a comment.

Subsets are not the same things as elements, so although $\emptyset$ is a subset of both $A$ and its complement, whether or not it is an element depends on what exactly is contained in $A$. If it is an element of $A$ it will not be an element of $A$'s complement.

Related: Is empty set element of every set if it is subset of every set? and Is the null set a subset of every set?

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  • $\begingroup$ Suppose A = { 1,2,3} and U = {1,2,3,4}. So A' = {4}. So will then A intersection A' = null? If yes, why? Null is not an element of any set here. $\endgroup$ – Ram Keswani Apr 4 '18 at 10:40
  • $\begingroup$ I would rather say: subsets of some set $A$ are not the same things as elements of $A$. This because in set-theory subsets and elements are both sets (so in that sense the same thing). $\endgroup$ – drhab Apr 4 '18 at 10:41
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    $\begingroup$ @RamKeswani If you are saying that $A\cap A^{\complement}=\varnothing$ then you are not saying at all that $\varnothing\in A\cap A^{\complement}$. The only thing you are saying is that $A\cap A^{\complement}$ has no elements. $\endgroup$ – drhab Apr 4 '18 at 10:44

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